Answer
$$\approx 0.21 \mathrm{W}$$
Work Step by Step
Using Eq. $18-32,$ the rate of energy flow through the surface is
$$P_{\text {cond }}=\frac{k A\left(T_{s}-T_{w}\right)}{L}$$$$=(0.026 \mathrm{W} / \mathrm{m} \cdot \mathrm{K})\left(4.00 \times 10^{-6} \mathrm{m}^{2}\right) \frac{300^{\circ} \mathrm{C}-100^{\circ} \mathrm{C}}{1.0 \times 10^{-4} \mathrm{m}}$$$$=0.208 \mathrm{W} \approx 0.21 \mathrm{W}$$
